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If the solutions of p(x)=0 are -14 and 11, which function could be p?

A. p(x)=
x^(2)-3x-154

B. p(x)=
x^(2)-14x+11

C. p(x)=
x^(2)+14x+11

D. p(x)=
x^(2)+3x-154

User Mar Cial R
by
5.3k points

2 Answers

3 votes

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Solution :


x = - 14


x = 11

Thus ;


x+ 14 = 0


x - 11 = 0

Multiply above equations :


(x + 14)(x - 11) = 0


{x}^(2) + ( 14 - 11)x +( \: ( 14) * ( - 11) \: ) = 0 \\


{x}^(2) + 3x - 154 = 0

So ;


p(x) = {x}^(2) + 3x - 154

Thus the correct answer is (( D )) .

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User Sohilv
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5.2k points
3 votes

The quadratic function p(x) = x² + 3x - 154 accurately represents p(x) with solutions -14 and 11, distinguishing it from other options.

The solutions of the quadratic equation p(x) = 0 are the values of x for which the equation is true. Given that the solutions are -14 and 11, we can use these values to determine which function could represent p(x).

The factored form of a quadratic equation is p(x) = (x - r₁)(x - r₂), where r₁ and r₂ are the roots or solutions of the equation. For the given solutions, -14 and 11, the factored form would be p(x) = (x + 14)(x - 11).

Now, let's expand this expression to check which function matches:

p(x) = (x + 14)(x - 11) = x² + 3x - 154

Comparing this with the given options:

A. p(x) = x² - 3x - 154 - Incorrect

B. p(x) = x² - 14x + 11 - Incorrect

C. p(x) = x² + 14x + 11 - Incorrect

D. p(x) = x² + 3x - 154 - Correct

Therefore, the function that could represent p(x) with roots -14 and 11 is:

D. p(x) = x² + 3x - 154

User Turbosqel
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5.4k points